Questions and Answers #7 Matrices

Questions and Answers Sheet 7 Matrices Question #1 It can be shown that the algebraic multiplicity of an eigenvalue 𝜆 is always greater than or equal to the dimension of the eigenspace corresponding to 𝜆 . Find h in the matrix A below such that the eigenspace for \lambda=5λ=5 is two-dimensional: 5 −2 0 3 A= 0 0  0 0 6 −1 h 0  5 4  0 1 Answer: Row reduce the augment matrix for the equation (A-5I)x=0 0 −2 0 −2  0 0  0 0 6 −1 h 0 0 4 0 −4 0  0 −2 6 −1   0  0 0 h − 6 1 = 0  0 0 0 4   0  0 0 0 4 0  0 1 −3 0 0  0 −2 h 0 = 0  0 0 0 4   0  0 0 0 −4 0 0  0  0 For a two-dimension eigenspace, the system above needs two free variables. This happens if and only if h=6 0 0  0  0 1 −3 0 0 0 0 0 0 0 0 1 0 0 0  0  0 h=6 Question #2 For the matrices (a) find k such that Nul A is a subspace of 𝑅𝑘 , and (b) find k such that Col A is a subspace of 𝑅𝑘 4 5 −2 6 0 A=  1 1 0 1 0  Answer: a) In order to find Nul A, we must solve Ax=0 Since A is 2 × 5 matrix, it can multyplied from left only by vector with 5 rows. That means x is in 𝑅𝟝 , i.e. k=5 b) As we saw in part a) 𝑥 ∈ 𝑅𝟝 , so we know Col 𝐴 = 𝑦: 𝑦 = 𝐴𝑥 for some 𝑥 ∈ 𝑅𝟝 So we have a multiplication of 2 × 5 matrix A and 5 × 1 vector x. From definition of matrix muplication, y is 2 × 1 vector, i.e k=2 Result: a) 5, b) 2 Question #3 Write the solution set of the given homogeneous system in parametric vector form. 𝑥1 + 3𝑥2 − 5𝑥3 = 0 𝑥1 + 4𝑥2 − 8𝑥3 = 0 −3𝑥1 − 7𝑥2 + 9𝑥3 = 0 Answer: The given system is 𝑥1 + 3𝑥2 − 5𝑥3 = 0 𝑥1 + 4𝑥2 − 8𝑥3 = 0 −3𝑥1 − 7𝑥2 + 9𝑥3 = 0 The augmented matrix is  1 3 −5 0  1 4 −8 0    −3 −7 9 0 Reduce matrix to row echelon form   −3 −7 9   0 5 −5  3  2 0 −2 3   −3 −7 9  0 1 −3   0 0 0  0  R − 2 / 5R → R 2 3 0 3  3 / 5R2 → R2  0  0 1 R1 + 7 R2 → R1   0 0 −1/ 3R1 → R1  0 0  −3 −7 9 0  ~  0 1 −3 0   0 0 0 0  0 4 0 1 −3 0  0 0 0  Thus we get 𝑥1 + 4𝑥3 = 0 ⇒ 𝑥1 = −4𝑥3 𝑥2 − 3𝑥3 = 0 ⇒ 𝑥2 = 3𝑥3 Hence, the given homogeneous system in parametric vector form is  x1   −4   x =  x2  = x3  3   x3   1  Question #4 Find a least-squares solution of Ax=b by (a) constructing the normal equations for 𝑥̂ and (b) solving for 𝑥̂  2 1  −5   A =  −2 0 , b =  8   2 3  1  Answer: We know that each least square solution of Ax=b satisfies the equation 𝐴𝑇 𝐴𝑥 = 𝐴𝑇 𝑏. Therefore first of all find 𝐴𝑇 𝐴, 𝐴𝑇 𝑏 2  2 −2 2   A A=   −2 1 0 3   2   2 −2 2  [ AT b =    −5 1 0 3  T 1 12 8  0  =  ] 8 10   3  −24  8 1 =    −2  The equation 𝐴𝑇 𝐴𝑥 = 𝐴𝑇 𝑏 becomes. 12 8   x1   −24  8 10   x  =  −2    2   The augmented matrix for 𝐴𝑇 𝐴𝑥 = 𝐴𝑇 𝑏 to solve this sytem 12 8 −24  8 10 −2    Use row operation to solve this system 12 8 −24 1 0 −4  8 10 −2  ~ 0 1 3      The solution of the augmented matrix. 𝑥1 = −4, 𝑥2 = 3 The general least square solution for Ax=b  −4 xˆ =   3 Question #5 2 3 Let y =   and 4 u=   −7  Write y as the sum of two orthogonal vectors, one in Span{u} and one orthogonal to u. Answer: Formula for orthogonal projection: 𝑦⋅𝑢 𝑦̂ = 𝑢 𝑢⋅𝑢 𝑦 ⋅ 𝑢 = (2,3) ⋅ (4, −7) = 8 − 21 = −13 𝑢 ⋅ 𝑢 = (4, −7) ⋅ (4, −7) = 16 + 49 = 65 yˆ = −13 −1  4   −4 / 5 u=  = 65 5 −7   7 / 5  𝑦̂ is the part that is in span u. Now calculate other part:  2   −4 / 5 14 / 5 y − yˆ =   −  = ]  3  7 / 5   8 / 5   −4 / 5 14 / 5 [y =  +   7/5   8/5  Question #6 Let 𝑅1 and 𝑅2 be a relations on a set A represented by the matrices below: 0 M R1 = 1 1 0 [ M R 2 = 0 1 1 0 1 1  ] 0 0  1 0 1 1  1 1  Find the matrix that represents 𝑅1 ∩ 𝑅2 Answer: 0 1 0  0 1 0      Consider the given matrices, M R1 = 1 1 1 and M R 2 = 0 1 1     1 0 0 1 1 1  Note that, the intersection 𝑎 ∩ 𝐵 is all elements that are both in A and in B. The matrix corresponding to the intersection of two relations is the meet of the matrices representing each of the relations. 𝑀𝑅1∩𝑅2 = 𝑀𝑅1 ∩ 𝑀𝑅2 0  0 = 1  0  1  1 0 1 = 0 1 1 0 1  1 0  0 1  1 1  1  0  1 0  1  1 1  0  Thus, the matrix that represents 0 1 1  R1  R2is = 0 1 1  1 0 0 Question #7 Use the given of the coefficient matrix to solve the following system. 7𝑥1 + 3𝑥2 = 6 −6𝑥1 − 3𝑥2 = 4 1  1   A =  −2 − 7  3  −1 Answer: 7𝑥1 + 3𝑥2 = 6 −6𝑥1 − 3𝑥2 = 4 Ax=B ⇒ 𝑥 = 𝐴−1 𝐵 7 3 6 A= B=    −6 −3  4 1  1 −1   6 x=A B= 7  −2 −   4  3  Here 7𝑥1 + 3𝑥2 = 6 −6𝑥1 − 3𝑥2 = 4 Ax=B ⇒ 𝑥 = 𝐴−1 𝐵 Here 7 3 6  A= B =  ]   −6 −3 4 1  1   6] [= 7  −2 −   4  3   6+4  ] [=  12 − 28  3   10  [=    −64 / 3 Question #8 1  −6  5       Let a1 = 5 , a2 = −26 ], and b = 9 . For what value(s) of h is b in the plane spanned by 𝑎1 and        2   −1  h  𝑎2 ? Answer: Given: 1  −6  5     a1 =  5  , a2 =  −26 , and b = 9   −1  2   h  To find value of h if b spanned by 𝑎1 and 𝑎2 Since b is spanned by 𝑎1 and 𝑎2 𝛼(1,5, −1) + 𝛽(−6, −26,2) = (5,9, ℎ) 𝛼 − 6𝛽 = 9 5𝛼 − 26𝛽 = 9 −𝛼 + 2𝛽 = ℎ Solving first two equation, we get 𝛼 = −19, 𝛽 = −4 ℎ = 11 Question #9 What are the values of these sums, where S={1,3,5,7}? a) ∑𝑗∈𝑆 𝑗 b) ∑𝑗∈𝑆 𝑗 2 c) ∑𝑗∈𝑆(1/𝑗) d) ∑𝑗∈𝑆 1 Answer: Step 1 Compute the sum of this expression by substituting the values from the set S and adding the together. PART A: ∑(𝑗)𝑗 ∈ 𝑆 = 1 + 3 + 5 + 7 = 16 Step 2 Compute the sum of this expression by substituting the values from the set S and adding the together. PART B: ∑(𝑗 2 )𝑗 ∈ 𝑆 = 12 + 32 + 52 + 72 = 84 Question #10 Find a basis for the space of 2 × 2 lower triangular matrices Basis = Answer: S = set of lower triangular matrices. To find basis of s a 0 A=  S] b c  Let 1 0  0 0  0 0  [A = a  +b +c    0 0  1 0  0 1  ∴ A can be written as the linear combination of vectors 1 0   0 0  0 0  1 0   0 0   0 0  0 0 , 1 0 and 0 1 and B = 0 0 , 1 0 , 0 1             is linearly independent set ∴ B is basis for the space of 2 × 2 lower triangular matrices Finally answer: 1 0   0 0   0 0  B= , ,   0 0  1 0   0 1  Question #11 Show that if A and B are similar matrices, then det(𝐴) = det(𝐵). Answer: Let A and B be similar. Then there exist a nonsingular matrix S such that 𝐴 = 𝑆 −1 𝐵𝑆. We have that det(𝐴) = det(𝑆 −1 𝐵𝑆) = det(𝑆 −1 ) det(𝐵) det(𝑆) = det(𝑆 −1 ) det(𝑆) det(𝐵) = det(𝑆 −1 𝑆) det(𝐵) = det(𝐼) det(𝐵) = det(𝐵). Result: Since A and B are similar, there exist a nonsigular S such that 𝐴 = 𝑆 −1 𝐵𝑆 . Then, using the fact that det(𝐴𝐵) = det(𝐴) det(𝐵) for any matrices A and B, the result is easily obtained. Question #12 Find the nth Taylor polynomial centered at c. 𝑓(𝑥) = 1/𝑥 2 , 𝑛 = 4, 𝑐 = 2 Answer: We need to find the 4th Taylor Polynomial for the function 𝑓(𝑥) = need to calculate 𝑓 ′ (−2.1), 𝑓 ′′ (−2.1), 𝑓 ′′′ (−2.1), 𝑓 𝑓(𝑥) = 𝑓 ′ (𝑥) 𝑓 ′′ (𝑥) 𝑓 ′′′ (𝑥) 𝑓 (4) (4) 1 𝑥2 centered at c=-2. Therefore, we (−2.1). 1 𝑥2 −2 = = 𝑥3 6 = 𝑥4 −24 (𝑥) = 𝑥5 120 𝑥6 1 1 −2 4 1 𝑓(−2.1) = (−2.1)2 = 𝑓 ′ (−2.1) = (−2.1)3 = 𝑓 ′′ (−2.1) 𝑓 ′′′ (−2.1) 𝑓 (4) 6 = (−2.1)4 = 4 3 −24 = (−2.1)5 = (−2.1) = 120 (−2.1)6 8 3 = 4 15 8 Therefore, the 4th Taylor Polynomial is 1 1 3 3 (𝑥 + 2)2 + (𝑥 + 2)3 + 𝑃4 (𝑥) = + (𝑥 + 2) + 8∗4! Result: 1 1 𝑃4 (𝑥) = + (𝑥 + 2) + 8∗4! 4 4 4 4 8∗2! 3 8∗2! 4∗3! (𝑥 + 2)2 + 3 4∗3! (𝑥 + 2)3 + 15 15 (𝑥 + 2)4 (𝑥 + 2)4 Question #13 Assume that T is a linear transformation. Find the standard matrix of T. 𝑇: 𝑅2 → 𝑅4 , 𝑇(𝑒1 ) = (3,1,3,1) and 𝑇(𝑒2 ) = (−5,2,0,0), where 𝑒1 = (1,0) and 𝑒2 = 90,1 ) Answer: We know, that 1  e1 =   0  0  e2 =   1  The task is given  3 1 T (e1 ) =    3   1  −5  2 T (e2 ) =   0   0 There exists a unique matrix A for the linear transformation T for which it holds T(u)=Au for all u and A is a form 𝐴 = [𝑇(𝑒1 ), … , 𝑇(𝑒𝑛 )], where 𝑒𝑖 , 𝑖 = 1,2, … are vectors from the identity matrix, respectively to columns. Matrix A is the standart matrix for the linear transformation T. So, the standart form is: 3 −5 1 2   A = [T (e1 )T (e2 )] =  3 0    1 0  3 −5 1 2   A= 3 0    1 0  Question #14 The nth order permutation matrix refers to an n × n matrix in which each column and row has exactly one 1 and the rest of the components are zero. Find the exponential generation function for the number of symmetric nth order permutations matrices. 1 0 0  0 0 1    0 1 0  Answer: Step 1 The series expansion of exponential function is given by 𝑒𝑥 = 1 + 𝑥 1! + 𝑥2 2! + 𝑥3 3! + ⋯ This series also converges for a square matrix A i.e. 𝑒𝐴 = 𝐼 + 𝐴 1! + 𝐴2 2! + 𝐴3 3! + ⋯ where I is the identity matrix of same order as A. Here given, 1 0 0  1 0 0  0 0 1  A2 = 0 0 1  = I     0 1 0 0 1 0 𝐴3 = 𝐴2 . 𝐴 = 𝐼. 𝐴 = 𝐴 𝐴4 = 𝐴2 . 𝐴2 = 𝐼. 𝐼 = 𝐼 so on Therefore we get the even power A is I where as odd power of A is A itself  I for k is even Ak =   A for k is odd Step 2 So we can write the series expansion as, 𝑒𝐴 = 𝐼 + 𝐴 1! + 𝐴2 2! + 𝐴3 3! + ⋯ 1 0 0  1 0 0  1 0 0  1 0 1 1 1     e =  0 1 0  +  0 0 1  + 0 1 0  + 0 0 1! 2! 3! 0 0 1  0 1 0  0 0 1  0 1  1 1 1 0 0 1 + 1! + 2! + 3! +  1 1 1 1 1 A e = 0 1+ + + + + +  2! 4! 1! 3! 5!  1 1 1 1 1  0 + + + 1+ + + 1! 3! 5! 2! 4!  A 0 1  + 0          By series expansion we know 𝑥 𝑥2 𝑥3 𝑥4 𝑒𝑥 = 1 + + + + + ⋯ 1! 2! 3! 4! 𝑒 𝑥 − 𝑒 −𝑥 𝑥3 𝑥5 𝑥7 sin ℎ 𝑥 = =𝑥+ + + +⋯ 2 3! 5! 7! 𝑒 𝑥 + 𝑒 −𝑥 𝑥2 𝑥4 𝑥6 cos ℎ 𝑥 = =1+ + + +⋯ 2 2! 4! 6! By putting x=1 1 1 1 𝑒1 = 1 + + + + ⋯ 2! 3! 4! 1 1 1 cos ℎ 1 = 1 + + + + ⋯ 2! 4! 6! 1 1 1 sin ℎ 1 = 1 + + + + ⋯ 3! 5! 7! Therefore we can express the require term as 0 0  1 + e  e =  0 cos h1 sin h1  0 sin h1 cos h1 A Question #15 (square roots of the identity matrix) For how many 2 × 2 matrices A is it true that 𝐴2 = 𝐼? Now answer the same quaestion for 𝑛 × 𝑛 matrices where n >2 Answer: Step 1 Consider a 2 × 2 matrix, 1   A=  0 −1 Now, 𝐴2 = 𝐴 ⋅ 𝐴 1    1    1 0  =  =  0 −1 0 −1 0 1 Hence all values 𝛼 the matrix satisfies. So there uncountable number of 𝛼 satisfying 𝐴2 = 𝐼. Step 2 Similarly for all n we get, 1  0  A = 0 −1 0  0 0 1  So, 𝐴2 = 𝐴 ⋅ 𝐴 1 = 0 0 1 = 0 0  0  1  0  −1 0  0 −1 0  0 1  0 0 1  0 0 1 0  0 1  =I For n=3 there exist uncountable number of 3 × 3 matrices A such that 𝐴2 = 𝐼 Thus the theorem satisfies for all 𝑛 × 𝑛 matrices. Question #16 Problem. Suppose that the 𝑛 × 𝑛 matrices A and B commute; that is, that AB = BA. Prove that 𝑒 𝐴+𝐵 = 𝑒 𝐴 𝑒 𝐵 . (Suggestion: Group the terms in the product of the two series on the right-hand side to obtain the series on the left.) Deduce from the result of Above problem that, for every square matrix A, the matrix 𝑒 𝐴 is nonsingular with (𝑒 𝐴 )−1 = 𝑒 −𝐴 Answer: Step 1 Consider 2 × 2 diagonal matrices A and B which commute,i.e AB=BA. a 0  a A= 1 ,B =  2   0 b1  0 0 b2  Then, a n An =  1 0 0  n  a2n ,B =  b1n  0 Step 2 For integer 𝑛 ≥ 1 We have, 𝑒𝐴 = 𝐼 + 𝐴 + 𝐴2 2! + ⋯ By putting the values, 0  b2n   a12 1 0   a1 0   2! eA =  + + 0 1   0 b1    0   a12 0 1 + a1 + 2!   = b12   0 1 + b1 +   2!  e a1 0  = b1  0 e   0 + b12  2!  So we can see that the exponential of the diagonal 2 × 2 matrix A is obtained by exponentiating each diagonal element of matrix A. Similarly for B, ea2 eB =  0 0  eb2  Step 3 e a1 0  e a2 0  e Ae B =  b1   b2   0 e  0 e  e a1 e a2 + 0 0  =  b1 b2 0 e e + 0  Now e a1 e a2 0  = b1 b2  e e   0 e a1 + a2 0  = b1 + b2  e  0  = 𝑒 𝐴+𝐵 Hence, 𝑒 𝐴 𝑒 𝐵 = 𝑒 𝐴+𝐵 Step 4  e a1 As e =  0 A we have 0  eb1  1 eb1 0    e a1 eb1  0 e a1   eb1  0   a1 b1 e e  =  e a1   0 e a1 eb1  1   ea1 0  e− a1 0  = =  (1) 1   0 e − b1  0  eb1   a12  0  1 0   a1 0  2! − e− A =  − +   b12  0 1   0 b1    0 2!  e − a1 0  = (2) − b1   0 e  (e A ) −1 = As (1) = (2) Hence (𝑒 𝐴 )−1 = 𝑒 −𝐴 Question #17 Find 4 × 4 matrices X and Y such that 0 1 XY − YX =  1  1 1 0 1 1 1 1 0 1 1 1  1  0 Answer: Step 1 a b Let XY =  b  b b a  c b and YX =  c b   a c 0 1 1 1 0 1 Then we will have XY − YX =  1 1 0  1 1 1 Step 2 b a b b b b a b c a c c 1 1  1  0 c c a c c c  such that 𝑎 ∈ 𝑅, b-a=1 and b-c=1 c  a Solving the linear equations b-a=1 and b-c=1, we get a=c. a b  XY =  b  b b a b b b a  a b − (1) and YX =  a b   a a b b a b a a a a a a a a a a  − (2) a  a Observe that, matrix (1) on the right side of the 1st equation has a full rank(determinant of it should be non zero), and matrix (2) has rank 1(determinant of it will be zero). Taking determinant on both sides of the above equations, we have det(𝑋) det(𝑌) = 𝑘 and det(𝑌) det(𝑋) = 0, where 𝑘 ∈ 𝑅 − 0 But this is impossible, as a product of two numbers cannot be zero and non-zero at the same time. Hence, there exist no such matrices X and Y such that 0 1 XY − YX =  1  1 1 0 1 1 1 1 0 1 1 1  1  0 Question #18 Find the least-square solution of Ax=b, where  1 −2  3  −1 2     and b =  1  A= 0 3  −4      2 5 2 Answer: Step 1 Given matrices:  1 −2  3  −1 2  1   A= and b =   0 3  −4      2 5 2 Now, Ax=b  1 −2  3  −1 2  x     1 =  1   0 3   x2   −4     2 5 2 Finding 𝐴𝑇 𝐴 Step 2  1 −1 0 2  AT A =    −2 2 3 5   1 −2   −1 2  −2 − 2 + 10    =  1+1+ 4   0 3   −2 − 2 + 10 4 + 4 + 9 + 25   2 5 6 6  =  6 42  Finding 𝐴𝑇 𝑏  1 −1 0 2  AT b =    −2 2 3 5  3 1    =  3 −1 + 4   −4   −6 + 2 − 12 + 10    2 6 =   −6  6 6   x1   6  thus,    =   6 42   x2   −6  Rewriting the above system in augmented matrix form and using reduced row echelon form, 6 6   6  1 1   1  R1 6 42   −6  R1  6  6 42   −6       1 1   1  R  R2  R2 − 6 R1   R2  2     36 0 36   −12   4   1  1 1   1 0   3    R  R − R  1 1 1 2  0 1   1  0 1  −    −   3  3  Step 3 thus, 𝑥1 = 4 3 and 𝑥2 = − 1 3 Question #19 A, B and C square matrices of order 4 and suppose that det(𝐴) = 2, det(𝐵) = 3 and det(𝐶) = 5. det(2𝐵𝐶 −1 𝐴𝑇 𝐶) Answer: Step 1 Given A, B and C square matrices of order 4 and suppose that det(A)=2, det(B)=3 and det(C)=5 we have to find det(2𝐵𝐶 −1 𝐴𝑇 𝐶) Step 2 Here, det(A)=2, det(B)=3 and det(C)=5 det(2𝐵𝐶 −1 𝐴𝑇 𝐶) = 2 det(𝐵) 𝑑𝑒𝑡(𝐶 −1 ) det(𝐴𝑇) det(𝐶) 1 we know that det (𝑋 −1 ) = and 𝑑𝑒𝑡(𝑋 𝑇 ) = 𝑑𝑒𝑡(𝑋) 𝑑𝑒𝑡(𝑋) 𝑑𝑒𝑡(2𝐵𝐶 −1 𝐴𝑇 𝐶) = 2𝑑𝑒𝑡(𝐵) 1 det ( A) det (C ) = 2det ( B)det ( A) det (C ) ⇒ 𝑑𝑒𝑡(2𝐵𝐶 −1 𝐴𝑇 𝐶) = 2𝑑𝑒𝑡(𝐵)𝑑𝑒𝑡(𝐴) = 2 × 3 × 2 = 12 hence, 𝑑𝑒𝑡(2𝐵𝐶 −1 𝐴𝑇 𝐶) = 12 this is the required answer Question #20 Solve the matrix equation for a, b, c, and d. 1 2  a b   6 3 3 4  c d  = 19 2      Answer: Step 1 The given condition of matrices are 1 2  a b   6 3 3 4  c d  = 19 2      Solving the matrices 1(a) + 2(c) 1(b) + 2(d )   6 3  a + 2c b + 2d   6 3 3(a) + 4(c) 3(b) + 4(d )  = 19 2 3a + 4c 3b + 4d  = 19 2        Equate both sides of above equation 𝑎 + 2𝑐 = 6 ⋯ (1) 𝑏 + 2𝑑 = 3 ⋯ (2) 3𝑎 + 4𝑐 = 19 ⋯ (3) 3𝑏 + 4𝑑 = 2 ⋯ (4) Step 2 To find the value of a,b,c and d, solve the above equations Now, perform the operation by subtracting 3 times of equation(1) from equation (3) 3a+4c=3(a+2c)=19-18 4c-6c=1 -2c=1 −1 𝑐= 2 Substitute the value of c in equation (1) −1 𝑎 + 2( ) = 6 2 a=6+1 =7 Solving equation (2) and equation (4) by subtracting 3 times of equation(1) from equation(4) 3b+4d-3(b+2d)=2-3(3) -2d=-7 7 𝑑= 2 Substituting the value of d in equation (2) 7 𝑏 + 2( ) = 3 2 b=3-7 =-4 Answer: Hence, the values of a,b,c and d are a=7 b=-4 −1 𝑐= 𝑑= 2 7 2 Question #21 A is a 2 × 2 matrix with elgenvectors 𝑣1 = [[1], [−1]] and 𝑣2 = [[1], [1]] corresponding to eigenvalues 𝜆1 = 1 2 and 𝜆2 = 2 , respectively,and 𝑥 = [[7], [1]], Find 𝐴8 𝑥 Answer: Step 1 1 GIven: 𝑉1 = ([1], [−1]); 𝑉2 = ([1], [1]) and 𝜆1 = ; 𝜆2 = 2 2 respectively Also given that: 𝑋 = ([7], [1]) To find: 𝐴8 𝑥 =? Step 2 To obtain: 𝐴8 =? So, 𝐴 = ([1,1], [−1,1]) ⇒ 𝐴2 = ([1,1], [−1,1])([1,1], [−1,1]) = ([0,2], [−2,0]) = 2([0,1], [−1,0]) 𝐴4 = 𝐴2 × 𝐴2 = 2([0,1], [−1,0]) × 2([0,1], [−1,0]) = 4([0,1], [−1,0]) × ([0,1], [−1,0]) = 4([−1,0], [0, −1]) Step 3 𝐴8 = 𝐴4 × 𝐴4 = 4([−1,0], [0, −1]) × 4([−1,0], [0, −1]) = 16([−1,0], [0, −1]) × ([−1,0], [0, −1]) = 16 = 16([1,0], [0,1]) = 16𝐼 Step 4 To obtain: 𝐴8 𝑋 =? 𝐴8 𝑋 = 16([1,0], [0,1])([7], [1]) = 16([7], [1]) = ([112], [16]) Step 5 Result: 𝐴8 𝑋 = ([112], [16])